Question

The electric field of a plane electromagnetic wave propagating along the $x-\text{direction}$ in vacuum is $\overrightarrow{E}=E_0\hat{j}\cos (\omega t-kx).$ The magnetic field $\overrightarrow{B},$ at the moment $t=0$ is:

• A. $\overrightarrow{B}=\frac{E_0}{\sqrt{{\mu }_0{\epsilon }_0}}\cos \left(kx\right)\hat{k}$
• B. $\overrightarrow{B}=E_0\sqrt{{\mu }_0{\epsilon }_0}\cos \left(kx\right)\hat{j}$
• C. $\overrightarrow{B}=E_0\sqrt{{\mu }_0{\epsilon }_0}\cos \left(kx\right)\hat{k}$
• D. $\overrightarrow{B}=\frac{E_0}{\sqrt{{\mu }_0{\epsilon }_0}}\cos \left(kx\right)\hat{j}$

Answer 1

The correct option is C $\overrightarrow{B}=E_0\sqrt{{\mu }_0{\epsilon }_0}\cos \left(kx\right)\hat{k}$

The relation between amplitude of electric field and magnetic field for an electromagnetic wave in vacuum is

$B_0=\frac{E_0}{c}$

In free space, its speed, $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$

Here, ${\mu }_0=$ absolute permeability, ${\epsilon }_0=$ absolute permittivity

$\therefore B_0=\frac{E_0}{c}=\frac{E_0}{1/\sqrt{{\mu }_0{\epsilon }_0}}=E_0\sqrt{{\mu }_0{\epsilon }_0}$

As the electromagnetic wave is propagating along $x-\text{direction}$ and electric field is along $y-\text{direction}$,

$\therefore \hat{E}\times \hat{B}||\hat{c}$
$\text{(Here,}\hat{c}\to \text{direction of propagation})$

$\therefore \overrightarrow{B}$ should be in $\hat{k}$ direction.

$\therefore B=E_0\sqrt{{\mu }_0{\epsilon }_0}\cos (\omega t-kx)\hat{k}$

At $t=0$
$B=E_0\sqrt{{\mu }_0{\epsilon }_0}\cos \left(kx\right)\hat{k}$

Hence, option $\left(C\right)$ is correct.